Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int n, m, q, cnt, root[maxn];
int deep[maxn], fa[maxn][32], flag[maxn];
vector<int> es[maxn], city[maxn];
void dfs(int u) {
  flag[u] = 1;
  for (auto v : es[u]) {
    if (flag[v]) continue;
    deep[v] = deep[u] + 1;
    fa[v][0] = u;
    dfs(v);
  }
}
int lca(int x, int y) {
  if (deep[x] > deep[y]) swap(x, y);
  for (int i = 31; i >= 0; i--)
    if (deep[fa[y][i]] >= deep[x]) y = fa[y][i];
  if (x == y) return x;
  for (int i = 30; i >= 0; i--)
    if (fa[x][i] != fa[y][i]) {
      x = fa[x][i];
      y = fa[y][i];
    }
  return fa[x][0];
}
void init(int n) {
  for (int i = 1; i <= n; i++) {
    if (!flag[i]) {
      deep[i] = 1;
      fa[i][0] = 0;
      dfs(i);
    }
  }
  for (int i = 1; i < 30; i++) {
    for (int j = 1; j <= n; j++) {
      fa[j][i] = fa[fa[j][i - 1]][i - 1];
    }
  }
}
struct Node {
  int ls, rs, sum;
} T[maxn * 40];
void upd(int l, int r, int& x, int y, int val) {
  T[++cnt] = T[y], x = cnt;
  if (l == r) {
    T[x].sum++;
    return;
  }
  int mid = l + r >> 1;
  if (mid >= val)
    upd(l, mid, T[x].ls, T[y].ls, val);
  else
    upd(mid + 1, r, T[x].rs, T[y].rs, val);
  T[x].sum = T[T[x].ls].sum + T[T[x].rs].sum;
}
int qr(int l, int r, int a, int b, int c, int d, int k) {
  if (l == r) return l;
  int mid = l + r >> 1;
  int sum = T[T[a].ls].sum + T[T[b].ls].sum - T[T[c].ls].sum - T[T[d].ls].sum;
  if (sum >= k)
    return qr(l, mid, T[a].ls, T[b].ls, T[c].ls, T[d].ls, k);
  else
    return qr(mid + 1, r, T[a].rs, T[b].rs, T[c].rs, T[d].rs, k - sum);
}
void dfs(int u, int fa) {
  if (city[u].size())
    upd(1, maxn, root[u], root[fa], city[u][0]);
  else
    T[root[u] = ++cnt] = T[root[fa]];
  for (int i = 1; i < city[u].size(); i++)
    upd(1, maxn, root[u], root[u], city[u][i]);
  for (auto v : es[u])
    if (v != fa) dfs(v, u);
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    es[u].push_back(v);
    es[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    int x;
    scanf("%d", &x);
    city[x].push_back(i);
  }
  init(n);
  dfs(1, 0);
  while (q--) {
    int u, v, a;
    scanf("%d%d%d", &u, &v, &a);
    int uv = lca(u, v);
    int k = min(a, T[root[u]].sum + T[root[v]].sum - T[root[uv]].sum -
                       T[root[fa[uv][0]]].sum);
    printf("%d ", k);
    for (int i = 1; i <= k; i++) {
      printf("%d ",
             qr(1, maxn, root[u], root[v], root[uv], root[fa[uv][0]], i));
    }
    puts("");
  }
}